Find the sum of consecutive terms of an arithmetic sequence Calculus

The sequence u_n = 3 + 2n is an arithmetic sequence. Therefore, its sum is:

S_n = \dfrac{n}{2}\left(f + l\right)

Where S_n is the sum of n terms in an arithmetic sequence, with f being the first term and l being the last term.

Here, there are a total of 8 terms summed, so:

n = 8

The first term and last term are:

• f = u_1 = 3 + 2\left(1\right) = 5
• l = u_8 = 3 + 2\left(8\right) = 19

Therefore:

\sum_{k=1}^{8} u_k = \dfrac{8}{2}\left(5 + 19\right) = 4\left(24\right) = 96

The sequence u_n = \dfrac{-n}{4} + \dfrac{1}{2} is an arithmetic sequence. Therefore, its sum is:

S_n = \dfrac{n}{2}\left(f + l\right)

Where S_n is the sum of n terms in an arithmetic sequence, with f being the first term and l being the last term.

Here, there are a total of 8 terms summed, so:

n = 8

The first term and last term are:

• f = u_1 = \dfrac{-1}{4} + \dfrac{1}{2} = \dfrac{1}{4}
• l = u_8 = \dfrac{-8}{4} + \dfrac{1}{2} = \dfrac{-3}{2}

Therefore:

\sum_{k=1}^{8} u_k = \dfrac{8}{2}\left(\dfrac{1}{4} + \dfrac{-3}{2}\right) = 4\left(\dfrac{-5}{4}\right) = -5

The sequence u_n = \dfrac{3n + 8n^2}{2n} is an arithmetic sequence. To realize this, simpfify the expression:

\dfrac{3n + 8n^2}{2n} = \dfrac{3n}{2n} + \dfrac{8n^2}{2n} = \dfrac{3}{2} + 4n

Therefore its sum is:

S_n = \dfrac{n}{2}\left(f + l\right)

Where S_n is the sum of n terms in an arithmetic sequence, with f being the first term and l being the last term.

Here, there are a total of 11 terms summed, so:

n = 11

The first term and last term are:

• f = u_5 = 4\left(5\right) + \dfrac{3}{2} = \dfrac{43}{2}
• l = u_{15} = 4\left(15\right) + \dfrac{3}{2} = \dfrac{123}{2}

Therefore:

\sum_{k=5}^{15} u_k = \dfrac{11}{2}\left(\dfrac{43}{2} + \dfrac{123}{2}\right)=\dfrac{11}{2}\left(\dfrac{166}{2}\right)= \dfrac{11}{2}\left(83\right)= \dfrac{913}{2}

The sequence u_n = 4n - 7 is an arithmetic sequence. Therefore, its sum is:

S_n = \dfrac{n}{2}\left(f + l\right)

Where S_n is the sum of n terms in an arithmetic sequence, with f being the first term and l being the last term.

Here, there are a total of 6 terms summed, so:

n = 6

The first term and last term are:

• f = u_3 = 4\left(3\right) - 7 = 5
• l = u_8 = 4\left(8\right) - 7 = 25

Therefore:

\sum_{k=3}^{8} u_k = \dfrac{6}{2}\left(5 + 25\right) = 3\left(30\right) = 90

The sequence u_n = \dfrac{5n}{2} + 2n + 6 is an arithmetic sequence. Therefore, its sum is:

S_n = \dfrac{n}{2}\left(f + l\right)

Where S_n is the sum of n terms in an arithmetic sequence, with f being the first term and l being the last term.

Here, there are a total of 16 terms summed, so:

n = 16

The first term and last term are:

• f = u_1 = \dfrac{5\left(1\right)}{2} + 2\left(1\right) + 6 = \dfrac{21}{2}
• l = u_{16} = \dfrac{5\left(16\right)}{2} + 2\left(16\right) + 6 = 40 + 32 + 6 = 78

Therefore:

\sum_{k=1}^{16} u_k = \dfrac{16}{2}\left(\dfrac{21}{2} + 78\right) = 8\left(\dfrac{177}{2}\right) = 4\left(177\right) = 708

The sequence u_n = 16 - \dfrac{4}{7}n is an arithmetic sequence. Therefore, its sum is:

S_n = \dfrac{n}{2}\left(f + l\right)

Where S_n is the sum of n terms in an arithmetic sequence, with f being the first term and l being the last term.

Here, there are a total of 43 terms summed, so:

n = 43

The first term and last term are:

• f = u_7 = 16 - \dfrac{4}{7}\left(7\right) = 16 - 4 = 12
• l = u_{49} = 16 - \dfrac{4}{7}\left(49\right) = 16 - 28 = -12

Therefore:

\sum_{k=7}^{49} u_k = \dfrac{43}{2}\left(12 - 12\right) = 0

The sequence u_n = -100 + 5n is an arithmetic sequence. Therefore, its sum is:

S_n = \dfrac{n}{2}\left(f + l\right)

Where S_n is the sum of n terms in an arithmetic sequence, with f being the first term and l being the last term.

Here, there are a total of 20 terms summed, so:

n = 20

The first term and last term are:

• f = u_0 = -100 + 5\left(0\right) = -100
• l = u_{19} = -100 + 5\left(19\right) = -100 + 95 = -5

Therefore:

\sum_{k=0}^{19} u_k = \dfrac{20}{2}\left(-100 - 5\right) = 10\left(-105\right) = -1\ 050